Question

The trigonometric equation ${\sin }^{-1}x=2{\sin }^{-1}a$ has a solution for

• A. $allrealvaluesofa$
• B. $|a|<\frac{1}{2}$
• C. $|a|\ge \frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}\le |a|\le \frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $|a|\ge \frac{1}{\sqrt{2}}$

Given, ${\sin }^{-1}\left(x\right)=2{\sin }^{-1}\left(a\right)$

$x=\sin \left(2{\sin }^{-1}\right(a\left)\right)$

$x=2\sin \left({\sin }^{-1}\right(a\left)\right)\cos \left({\cos }^{-1}\right(\sqrt{1-a^2}\left)\right)$

$\therefore x=2a\sqrt{1-a^2}$

Now $xϵ[-1,1]$

Therefore
$|2a\sqrt{1-a^2}|\le 1$

$4a^2(1-a^2)\le 1$

$4a^4-4a^2+1\ge 0$

Hence,
$(2a^2-1{)}^2\ge 0$

$2a^2-1\ge 0$

$a^2\ge \frac{1}{2}$

$|a|\ge \frac{1}{\sqrt{2}}$

And $|a|$ has to be less than $1$

$a\in [-1,\frac{-1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},1]$